How can i resolve this equation?

Question

Consider the following property $ P(n) $:

$ \sum_{k=1}^{n} k = \frac{1}{8}(2n+1)^2 $

Show that

$\forall n (P(n) \Longrightarrow P(n+1))$

Where do i start?

Answer 1

You want to start by breaking it down, so: $P(n)=\sum_{k=1}^n\frac{1}{8}(2k+1)^2=\frac{1}{2}\sum_{k=1}^nk^2+\frac{1}{4}\sum_{k=1}^nk+\frac{1}{8}\sum_{k=1}^n1=\frac{1}{2}\sum_{k=1}^nk^2+\frac{1}{4}\sum_{k=1}^nk+\frac{1}{8}n$.

Now use these summation formulas:

$\sum_{k=1}^nk=\frac{n(n+1)}{2}$, and $\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$ to argue by induction.

Answer 2

The following is always true for all $n \geq 1$: $$\sum_{k=1}^nk = \frac{n(n+1)}{2}$$

You are looking for those $n$ such that:

$$\frac{n(n+1)}{2} = \frac{(2n+1)^2}{8}$$ $$4n(n+1) = (2n+1)^2$$ $$4n^2 + 4n = 4n^2 + 4n + 1$$ $$0 = 1$$

This means that $\mathcal{P}(n)$ is always false for every $n$.

Then $$\forall n(\mathcal{P}(n) \Rightarrow \mathcal{P}(n+1))$$ is true, since

$$A \Rightarrow B = \bar{A} \vee B$$

If $A$ is false ($\mathcal{P}(n)$), then $\bar{A}$ is true and hence $\bar{A} \vee B$ is true.